51Nod 1043 幸运号码

 #include <stdio.h>
 #include <algorithm>
 using namespace std;

 typedef long long ll;
 const int mod = 1e9 + ;
 int dp[][];
 // dp[i][j] : i 个数，组成总和为j 的数量

 int main()
 {
     int n;
     scanf("%d", &n);
     dp[][] = ;
     for (int i = ; i <= ; i++)
         dp[][i] = ;
     for (int i = ; i <= n; i++)
     {
         for (int j = ; j <=  * i; j++)
         {
             int sum = ;
             for (int k = ; k <= ; k++)
             {
                 if (j >= k)
                     sum = (sum + dp[i - ][j - k]) % mod;
                 else
                     break;
             }
             dp[i][j] = sum;
         }
     }
     ll ans = ;
     for (int i = ; i <=  * n; i++)
         ans = (ans + (ll)dp[n][i] * (dp[n][i] - dp[n - ][i])) % mod;
     printf("%lld\n", ans);
     return ;
 }